Question

X =x² + 6x + 5 = 0b= ] ± √b22-4aaC

Answer 1

We have the following quadratic equation:

$x^2+6x+5=0$

And we have to use the quadratic equation to solve that equation.

1. To find the solutions, we need to start by using the quadratic equation:

$\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ \text{ when }ax^2+bx+c=0 \end{gathered}$

2. Identify a, b, and c from the given quadratic equation:

$\begin{gathered} x^2+6x+5=0 \\ \\ a=1,b=6,c=5 \end{gathered}$

3. Apply the quadratic equation:

$\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-6\pm\sqrt{6^2-4(1)(5)}}{2(1)} \\ \\ \\ \end{gathered}$

4. And now, we can develop it as follows:

$\begin{gathered} x=\frac{-6\pm\sqrt{36-20}}{2} \\ \\ x=\frac{-6\pm\sqrt{16}}{2} \\ \\ x=\frac{-6\pm4}{2} \\ \\ \text{ Therefore:} \\ \\ x=\frac{-6+4}{2}=-\frac{2}{2}=-1 \\ \\ x=-1 \\ \\ x=\frac{-6-4}{2}=\frac{-10}{2}=-5 \\ \\ x=-5 \end{gathered}$

Therefore, we finally have two solutions, x = -1, and x = -5.

Then the first step is given by:

$x=\frac{-6\pm\sqrt{6^2-4(1)(5)}}{2(1)}$